The axiom of choice is a common set-theoretic axiom with many equivalents and consequences. This tag is for questions on where we use it in certain proofs, and how things would work without the assumption of this axiom.

learn more… | top users | synonyms

0
votes
1answer
47 views

Why is the statement “all vector space have a basis” is equivalent to the axiom of choice? [duplicate]

I'm reading a section in an abstract algebra book, where it reviews vector spaces and suddenly comments that "all vector space have a basis" is equivalent to the axiom of choice...I haven't studied ...
2
votes
1answer
38 views

Ordinal enumeration in ordered Mostowski model - does it not need the global choice?

In Jech's axiom of choice he proves following lemma (lemma 4.5(b) in his book): There is a injective mapping from $M$ to $\mathrm{Ord}\times \operatorname{fin}(A)$, where $A$ is a set of atoms and ...
0
votes
1answer
17 views

Show that the sets of isolated point of $E$ is a countable set - Axiome of choice

Let $E \subset \mathbb{C}$. Show that the sets of isolated point of $E$ is a countable set. That question is related to this question. However, my question somewhat different. Define ...
2
votes
2answers
61 views

Does the following condition implies full outer measure?

Let $X \subseteq 2^{\omega}$ be a set of positive Lebesgue measure. Suppose that for every $\eta, \nu \in 2^{<\omega}$ of the same length, the measure of $X$ above $\eta$ is the same as the measure ...
0
votes
3answers
43 views

What does Axiom of Choice mean [duplicate]

So this a fundamental assumption in mathematics. Can someone explain informally what it actually is please. My guess is that its when we say in proofs that "Let $x \in X$". But I am not sure.
0
votes
1answer
31 views

Prob. 9, Sec. 19 in Munkres' TOPOLOGY, 2nd edition: Equivalence of the choice axiom and non-emptyness of Cartesian product

The Axiom of Choice is as follows: Given a collection $\mathcal{A}$ of disjoint non-empty sets, there exists a set $C$ consisting of exactly one element from each element of $\mathcal{A}$; that ...
6
votes
1answer
102 views

Well-foundedness of cardinals and the axiom of choice

Without axiom of choice it is not generally true that the class of all cardinal (in this question we consider Scott cardinal rather than cardinals as ordinals) is not well-founded under the ordinary ...
4
votes
1answer
56 views

Algebraic closure with no nontrival automorphism

In Milne's notes on Galois theory, Chapter 7, p.91 he remarked that it is consistent without the axiom of choice that there exists an algebraic closure $L$ of $\mathbb{Q}$ with no nontrivial ...
1
vote
1answer
87 views

Proof of the Axiom of Choice

This exercise is from Bloch's book and can be found here. Bloch introduces equivalent variations of the axiom of choice where the one that will be proven is stated in terms of functions: AC1 Let ...
3
votes
1answer
79 views

Is there a choice homomorphism?

Let $\pi : \mathbb{R} \to \mathbb{R}/ \mathbb{Q}$ be the canonical projection. With the axiom of choice we "know" that there are choice functions $\alpha : \mathbb{R}/ \mathbb{Q} \to \mathbb{R}$ with ...
2
votes
1answer
44 views

Different Zorn's lemma statements

Given a chain-complete poset $P$, every $x\in P$ lies below some maximal element. Every inductive poset has enough maximal elements a maximal element. Chain-complete means every chain has a least ...
1
vote
1answer
33 views

Surjective function from a countable set

In Lang "Real and Functional analysis" is demonstrated that given a countable set $A$ and a function $f: A \rightarrow B$ which is surjective on $B$, then $B$ is finite or countable. Proof: Consider ...
3
votes
2answers
199 views

Hahn-Banach Theorem for separable spaces without Zorn's Lemma

I was reading about the Hahn-Banach Theorem, its many versions and their proofs. It's known that in the proofs we need Zorn's Lemma. But in the book that I'm reading, the author said if $X$ is a ...
1
vote
2answers
57 views

How does any map have a “pseudosection” (assuming axiom of choice)?

In the Lawvere and Rosebrugh book, Sets for Mathematics, exercise 4.34 is to show that the following is equivalent to the axiom of choice (every epimap has a section aka right inverse): If $f:X\to ...
2
votes
1answer
102 views

Can a large $V_\alpha$ satisfy Comparability but not be well-orderable?

Say that a set satisfies Comparability if any two of its subsets are comparable: one is injectable into the other. Are there models of ZF containing ranks $V_\alpha$ which satisfy Comparability but ...
0
votes
0answers
16 views

Is this Lindeloff theorem using AC? [duplicate]

Theorem: the following are equivalent: 1) The metric space $X$ is separable. 2) $X$ is second-countable (it has a countable basis) proof: $1 \Rightarrow 2: \lbrace B(d,r) : d\in D, r \in ...
1
vote
1answer
41 views

Lebesgue-measurable sets requiring the Axiom of Choice to construct

Every known construction of the Vitali set relies on the Axiom of Choice. It happens to not be Lebesgue-measurable. Must every set whose construction relies on the Axiom of Choice not be ...
2
votes
1answer
34 views

Alternate definitions of width (of a partial order) without Choice?

Say an antichain of a poset $P$ is a set of pairwise incomparable elements of $P.$ Typically, the width of a partial order is defined to be the supremum of the cardinalities of antichains of $P.$ When ...
6
votes
1answer
81 views

How can one prove the axiom of collection in ZFC without using the axiom of foundation?

Say I want to prove the axiom(s) of collection from the axiom(s) of replacement. If you have the axiom of foundation, then you can use Scott's trick to do this. But suppose I'm working in a context ...
8
votes
3answers
873 views

The smallest infinity and the axiom of choice

The short version of this question is: which (natural) axiom should be added to ZF so that the statement "$\aleph_{0}$ is the smallest infinity" becomes true? A set $A$ is called infinite if it can ...
1
vote
0answers
50 views

$|A^2|=|A|$ for every infinite $A$ iff Axiom of Choice holds. [duplicate]

I've seen this assertion in a few comments around the site, and I found the answer to the $\rightarrow$ implication here. Does anyone know a (hopefully simple) proof of the $\leftarrow$ implication?
1
vote
1answer
96 views

What is the first order formulation of Zorn's lemma in the language of set theory?

Very often in notes of courses in set theory you find the assertion that in ZF the Axiom of choice (AC) is equivalent to Zorn's Lemma (ZL) (which is equivalent to Well Ordering Principle which it ...
1
vote
0answers
43 views

Proving implication on well ordered set implies AC

Consider the following statement: If $A$ is a well-ordered set such that every nonempty subset of $A$ has a maximal element, then $A$ is finite. I am trying to prove that this statement implies ...
4
votes
1answer
65 views

Countable cartesian product and Axiom of Choice

In the A taste of Topology book, when talking about Cartesian product $\prod\{S:S\in\mathcal{S}\}$, the author writes the following: It is straightforward that ...
1
vote
3answers
168 views

Ordinal with given cardinality (without AC)

Is it possible to show that every cardinality has an ordinal with this cardinality (without the axiom of choice)? If so, how?
1
vote
0answers
52 views

Proof: Every lattice has a maximal filter iff AC

I'm working through a proof of Herrlich's book Axiom of Choice, p.58 (Google books): Equivalent are Every lattice has a maximal filter. Axiom of Choice. In this book, a lattice is ...
7
votes
3answers
196 views

Existence of mathematical objets constructed using the axiom of choice

Let consider the Vitali set $V \subset \mathbb R$, which is constructed using the axiom of choice. (I could take any other mathematical "object" that can be constructed using the axiom of choice, but ...
2
votes
0answers
37 views

How Does the Following Definition of the Axiom of Choise Entail that Elements Are Simultaneously Chosen from an Infinite Collection of Nonempty Sets [duplicate]

The following excerpt is from Ethan Bloch, Proofs and Fundamentals: A First Course in Abstract Mathematics (2nd ed, 2011 : page 121) and concerns one of the motivations for introducing the axiom of ...
5
votes
2answers
106 views

Does the law of the excluded middle imply the existence of “intangibles”?

First off, I'm not sure if "intangible" is standard terminology, Wikipedia defines an intangible object to be: "objects that are proved to exist, but which cannot be explicitly constructed". So if ...
1
vote
1answer
41 views

Axiom Of Choice to create a sequence of right inverses

I want to construct a sort of sequence of right inverses. My question is whether the construction uses the Axiom Of Choice correctly. Suppose I have a sequence of surjective functions $$ ...
8
votes
2answers
164 views

What's an example of a vector space that doesn't have a basis if we don't accept Choice?

I've read that the fact that all vector spaces have a basis is dependent on the axiom of choice, I'd like to see an example of a vector space that doesn't have a basis if we don't accept AoC. I'm ...
0
votes
0answers
32 views

Prob 2(d) Sec 9 in Munkres' TOPOLOGY 2nd edition: Is it possible to construct a choice function? [duplicate]

Here's Lemma 9.2 in Topology by James R. Munkres, 2nd edition: Given a collection $\mathscr{B}$ of non-empty sets (not necessarily disjoint), there exists a function $$c \colon \mathscr{B} \to ...
8
votes
2answers
275 views

Axiom of Choice needed to “categorify” the cardinals?

I was playing around in $\mathsf{Set},$ trying to reduce it modulo isomorphisms to make a category $\mathsf{Card},$ letting the objects of $\mathsf{Card}$ be the isomorphism classes of $\mathsf{Set}$ ...
5
votes
1answer
76 views

Is it consistent that every set is the countable union of sets with smaller cardinality, or is it just alephs?

(note: in what follows by "consistent" I mean "consistent relative to large cardinals") My question regards the exact statement of result which Gitik has proven in his paper "All Uncountable ...
4
votes
1answer
160 views

Existence of Hamel basis, choice and regularity

Blass (1984) shows that the existence of Hamel basis for arbitrary vector space over any field implies the axiom of choice. However such implication needs the axiom of regularity. As in Blass' ...
0
votes
2answers
71 views

Difference Between Axiom of choice and axiom of countable choice.

My question is: In particular, does the result that every surjective (continuous or even linear if it matters) function has a pre-inverse depend on the full axiom of choice or just the axiom of ...
0
votes
0answers
33 views

In $ZF$, $AC$ is equivalent to $\forall \alpha (\mathscr P(\alpha)$ can be well-ordered) [duplicate]

This is an exercise from Kunen - An introduction to independence proofs that I have hard time to solve. In $ZF$, $AC$ is equivalent to $\forall \alpha (\mathscr ...
4
votes
1answer
80 views

$\neg \textsf{AC}+ \neg\textsf{CH}$

Is there some interesting\surprising results that have only been proven by assuming $\neg \textsf{AC}$ and $\neg\textsf{CH}$ ? Is there some interesting\surprising results implying both $\neg ...
2
votes
6answers
113 views

What is some simple to prove very counter-intuitive result obtained by Choice?

I'm aware of some theorems like the Banach-Tarski's which yield very counter-intuitive results, however, it's proof is far beyond my knowledge, so I'm looking for some result that is easy to prove ...
6
votes
1answer
85 views

What is a projective object in $\rm Set$?

What property of a set in $\sf{ZF}$ is equivalent to its being a projective object in the category $\rm Set$? Since all sets are projective assuming $\sf AC$ my guess is that it is equivalent to ...
2
votes
2answers
36 views

Representing a vector space as a sum of subspaces

In a linear algebra book I'm reading now, there was the following exercise: Let $W\subseteq V$ be a subspace of vector space $V$. Do there always exist two subspaces $W_1,W_2\subseteq V$ such ...
1
vote
0answers
36 views

On Cantor's argument [duplicate]

What axioms of set theory are needed for Cantor's diagonalization argument to work and why? What happens if we do away with some of these axioms (for instance Axiom of Choice)?
1
vote
2answers
71 views

Proving the Axiom of Choice for countable sets

I am new to the axiom of choice, and currently working my way through some exercises. I am struggling with the following exercise: Exercise - Prove the Axiom of Choice (every surjective $f: X \to Y$ ...
3
votes
2answers
66 views

Does this proof make use of the Axiom of Choice?

Theorem Let $X$ be uncountable. Let $A$ be countable. Then $|X\cup A| = |X|$. Proof As $|X|>\aleph_0 \rightarrow \exists f: \Bbb N \to X$ injective. Note that $\operatorname ...
8
votes
1answer
114 views

Do the ZF-provable forcing principles differ from the ZFC-provable forcing principles?

In "The Modal Logic of Forcing", Joel David Hamkins and Benedikt Löwe show that the ZFC-provable forcing principles are exactly those of the modal logic S4.2 (interpreting $\Diamond \phi$ as asserting ...
0
votes
2answers
51 views

Axiom of choice and the empty set

Could someone explain to me why it is important for a set to be non-empty when working with a choice function?
3
votes
1answer
88 views

Why isn't there a total order of $\cal P(\Bbb R)$?

I have heard that, in some models of ZF, $\cal P(\Bbb R)$ has no total order. How could one prove this? I already know that $\Bbb R$ isn't necessarily well-ordered. I'm guessing that one could reduce ...
1
vote
2answers
62 views

Context for Russell's Infinite Sock Pair Example

I wanted to verify the following considerations on the context of Russell's infinite sock pair conundrum. The conundrum pointed out that a rule for choosing from pairs of shoes is possible a-priori. ...
-1
votes
1answer
114 views

Axiom of choice in HoTT without sethood requirement

3.8.3 of the HoTT book gives the following as a variant of the axiom of choice: $\Pi$ (X : U) (Y : X $\rightarrow$ U), (isSet X) $\rightarrow$ ($\Pi$ (x : X), isSet (Y x)) $\rightarrow$ ($\Pi$ (x ...
6
votes
6answers
403 views

When using an axiom scheme, are we implicitly using a choice principle?

I heard an interesting argument from a colleague recently that went something like this. Whenever we are using an axiom scheme, we are essentially choosing one of the instances of this scheme, and ...